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I'm not sure if my logic holds up here. The 4 and 2 are not relevant, I am more interested in the general case. I'll present a very informal version of what I was thinking of doing.

$|x_n - 4| < \epsilon$, rewrite this as

$|(\sqrt{x_n} -2)(\sqrt{x_n} + 2)| < \epsilon$

$|(\sqrt{x_n} -2)||(\sqrt{x_n} + 2)| < \epsilon$

$|(\sqrt{x_n} -2)| < \frac{\epsilon}{|(\sqrt{x_n} + 2)|} < \epsilon$

$|(\sqrt{x_n} -2)| < \epsilon$

Its really that fourth step that I am not sure about.

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    What is $\epsilon (4)$ supposed to mean? Note $\sqrt{x_n} \ge 0$ so $|\sqrt{x_n} + 2| \ge 2$ s. $\frac{\epsilon}{|\sqrt{x_n} + 2|} \le \frac {\epsilon}2$ So let $\delta = 2\epsilon$ then $|x_n -4| < \delta \implies |\sqrt{x_n} - 2|< \epsilon$. Use that instead. – 2017-01-24
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    Oh I just wanted to mark the step that I was unsure about with the (4). – 2017-01-24
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    Also make sure to prove that there exists $N_0$ such that $x_n > 0$ for all $n>N_0$ – 2017-01-24

3 Answers 3

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Well, it follows that if $x_n>0$, then

$$\frac\epsilon{\sqrt{x_n}+2}<\frac\epsilon2<\epsilon$$

In general, you could try to show that

$$\frac1{a+b}<\frac1b$$

for $a,b>0$.

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    It doesn't need $b>0$. – 2017-01-24
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    Yes, it does... – 2017-01-24
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    @MyGlasses: Yes, we need to know $a,b>0$. Take $a=2$ and $b=-1$. Hopefully you would agree the final line doesn't remain true. – 2017-01-24
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    @Clayton Oh, right. But if we have $a+b$ and $b$ both having the same sign, then it follows regardless through monotonicity. :D Got it! – 2017-01-24
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    @SimplyBeautifulArt: If we take $a=b=-1$, then $a+b$ and $b$ have the same sign, but we still don't have $\frac{1}{a+b}<\frac1b$. – 2017-01-24
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Since the function $f=\sqrt{x}$ is continuous at $x_{0}=4$, we have $f(x_{n})$ converges to $f(x_{0})$.

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I've actually found a better proof perhaps.

$|x_n - 4| = |\sqrt{x_n} - 2||\sqrt{x_n} + 2| < \epsilon$, it follows from $x_n \rightarrow 4$ that

$|x_n - 4| < \epsilon$, for $\epsilon = 1$ we have

$3 < x_n < 5$ and therefore $\sqrt{3} < \sqrt{x_n} < \sqrt{5}$ and we can see that

$\sqrt{x_n}$ is bounded from above by $\sqrt{5}$ so therefore

$|x_n - 4| = |\sqrt{x_n} - 2||\sqrt{x_n} + 2| < \sqrt{5}|\sqrt{x_n} + 2| < \epsilon$ thus

$|\sqrt{x_n} + 2| < \frac{\epsilon}{\sqrt{5}} < \epsilon$